Question

Two circles with equal radii are intersecting at the points (0, 1) and (0, -1). The tangent at the point (0, 1) to one of the circles. Then the distance between the centres of these circles is:
A. 1
B.\[\sqrt{2}\]
C. 2$\sqrt{2}$
D. 2

Answer 1

Hint: To solve the above question, first of all draw a diagram using the given information. Mark the points, to solve it easily. Use the equation $\sqrt{2}a$ to find the diagonal of a square having side-length ‘a’. Use the equation to find the hypotenuse of a right-angled triangle. That is, for right-angled $\vartriangle abc$, the equation for hypotenuse ‘ac’ is, ${{(ac)}^{2}}={{(ab)}^{2}}+{{(bc)}^{2}}$.

Complete step by step answer:

First of all we have to draw a diagram using the given information. It is given that there are two circles with equal radii.

So let us assume the radius of both the circles is ‘r’.

The intersecting points of the two circles are given as (0, 1) and (0, -1). Also given that the tangent at the point (0, 1) to one of the circles passes through the centre of the other circle.

Now, we can consider a square ACBD, so the distance between AB =$\sqrt{2}$r, as the length of the sides of the square is ‘r’.

So, the length of AO is $\dfrac{\sqrt{2}}{2}$r.

Here, CO = 1

Now, let us consider the$\vartriangle $ACO. So, now we have to use the equation of hypotenuse.

${{(AC)}^{2}}={{(CO)}^{2}}+{{(AO)}^{2}}$

On substituting the values, we get,

${{r}^{2}}={{1}^{2}}+{{\left( \dfrac{\sqrt{2}r}{2} \right)}^{2}}$

On solving the equation, we get,

${{r}^{2}}=1+\dfrac{2}{4}{{r}^{2}}$

$\Rightarrow {{r}^{2}}=1+\dfrac{1}{2}{{r}^{2}}$

Now, we can rearrange the equation,

\[ {{r}^{2}}-\dfrac{1}{2}{{r}^{2}}=1 \]

\[ \Rightarrow \dfrac{1}{2}{{r}^{2}}=1 \]

 \[ \Rightarrow {{r}^{2}}=2 \]

$\Rightarrow $r = $\pm \sqrt{2}$

Case 1: If we take r =$\sqrt{2}$, then,

$AO=\dfrac{\sqrt{2}}{2}r$

 So, distance between the centres of the circle,

AB = 2 X (AO)

$\Rightarrow $ $AB=2\times \dfrac{\sqrt{2}}{2}\times \sqrt{2}$

$\Rightarrow AB=2$

Case 2: If we take r =$-\sqrt{2}$, then,

$AB=2\times \dfrac{\sqrt{2}}{2}\times \left( -\sqrt{2} \right)$

So, AB = -2.

Therefore, we get the distance between the centres of the circle is 2.

So, the correct answer is “Option D”.

Note: We should be careful while drawing the diagram. Remember that the diagonal of a square with side r is$\sqrt{2}$r. When the value is either positive or negative, we have to substitute both the values to find the answer.